Unit Checkpoint

Answer: The Law of Segregation states that the two alleles for each gene separate during gamete formation, so each gamete receives only one allele.

Example: In a cross between two heterozygous tall pea plants (Tt x Tt), each parent's T and t alleles segregate during meiosis. Each gamete receives either T or t (not both). The offspring genotypes are TT, Tt, Tt, and tt in a 1:2:1 ratio, demonstrating that parental alleles separated and recombined in offspring.

Answer:

Leading Strand:

• Synthesized continuously in the 5' to 3' direction
• Runs toward the replication fork
• Requires only one RNA primer

Lagging Strand:

• Synthesized discontinuously in short Okazaki fragments
• Runs away from the replication fork
• Requires multiple RNA primers (one for each fragment)
• Requires DNA ligase to join fragments

Similarity: Both use DNA polymerase, synthesize in the 5' to 3' direction, and produce complementary strands.

Answer:

Punnett Square:

          X^C        X^c
X^C     X^C X^C    X^C X^c
Y       X^C Y      X^c Y
              

Genotypic Ratio: 1 X^C X^C : 1 X^C X^c : 1 X^C Y : 1 X^c Y

Phenotypic Ratio:

• Daughters: All normal vision (50% carrier, 50% non-carrier)
• Sons: 50% normal vision, 50% color blind

Answer: DNA replication is semi-conservative because each new double helix contains one original strand and one newly synthesized strand.

Meselson-Stahl Experiment (1958):

1. E. coli were grown in heavy nitrogen (15N) medium, making their DNA dense
2. Bacteria were transferred to light nitrogen (14N) medium
3. After one generation, DNA was intermediate density (one heavy, one light strand)
4. After two generations, half the DNA was intermediate, half was light

This pattern could only occur with semi-conservative replication, disproving conservative and dispersive models.

Answer: 1 Red : 2 Pink : 1 White

Explanation: This is incomplete dominance. Rr x Rr produces:

• RR (red) = 25%
• Rr (pink) = 50%
• rr (white) = 25%

Unlike complete dominance where the phenotypic ratio is 3:1, incomplete dominance shows a 1:2:1 phenotypic ratio matching the genotypic ratio.

Answer:

Autosomal Dominant:

• Trait appears in every generation
• Affected individuals have at least one affected parent
• Males and females equally affected

Autosomal Recessive:

• Trait can skip generations
• Two unaffected parents can have affected children
• Males and females equally affected

X-linked Recessive:

• More males affected than females
• Affected males have carrier mothers
• No father-to-son transmission
• All daughters of affected males are carriers

Answer:

• Helicase: Unwinds the double helix by breaking hydrogen bonds between base pairs; creates the replication fork
• Primase: Synthesizes short RNA primers that provide a 3'-OH group for DNA polymerase to begin synthesis
• DNA Polymerase: Adds nucleotides to the 3' end of the growing strand; reads template in 3' to 5' direction and synthesizes in 5' to 3' direction; also has proofreading ability
• Ligase: Joins Okazaki fragments on the lagging strand by creating phosphodiester bonds between fragments

Answer: 1/16

Explanation: YyRr x YyRr

For green and wrinkled, offspring must be yyrr (homozygous recessive for both traits)

• Probability of yy = 1/4 (from Yy x Yy)
• Probability of rr = 1/4 (from Rr x Rr)
• Probability of yyrr = 1/4 x 1/4 = 1/16

This matches the 9:3:3:1 ratio where the double recessive is 1 out of 16.

Answer: Mutations are changes in the DNA sequence that can alter the protein produced, potentially causing disease.

Sickle Cell Anemia Example:

• A single nucleotide substitution (A to T) in the hemoglobin gene changes the codon GAG to GTG
• This changes glutamic acid to valine at position 6 of the beta-globin chain
• The altered hemoglobin (HbS) causes red blood cells to become sickle-shaped under low oxygen conditions
• Sickle cells can block blood vessels, causing pain and organ damage

This demonstrates how a single base change can have severe phenotypic consequences.

Answer: Perform a test cross by crossing the tall plant with a homozygous recessive short plant (tt).

Expected Results:

If the tall plant is TT:

• TT x tt produces all Tt offspring
• All offspring will be tall
• 100% tall offspring indicates homozygous dominant parent

If the tall plant is Tt:

• Tt x tt produces 50% Tt and 50% tt
• Offspring will be 50% tall, 50% short
• Appearance of short offspring indicates heterozygous parent

A test cross uses a homozygous recessive individual to reveal the unknown genotype based on offspring phenotypes.